Sequence and SeriesMCQPYQ Dec. 22Question 1835 of 212
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In a GP 5th\displaystyle 5^{th} term is 27 and 8th\displaystyle 8^{th} term is 729. Find its 11th\displaystyle 11^{th} term?

Options

A729
B6561
C2187
D19683
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Correct Answer

Option d19683

All Options:

  • A729
  • B6561
  • C2187
  • D19683

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Detailed Solution & Explanation

Let the first term be a\displaystyle a and common ratio be r\displaystyle r.
Given:
t5=ar4=27t_5 = ar^4 = 27
t8=ar7=729t_8 = ar^7 = 729
Dividing:
r3=72927=27    r=3r^3 = \frac{729}{27} = 27 \implies r = 3

The 11th\displaystyle 11^{\text{th}} term is:
t11=ar10=ar7r3=72933=72927=19683t_{11} = ar^{10} = ar^7 \cdot r^3 = 729 \cdot 3^3 = 729 \cdot 27 = 19683
Hence, **Option D** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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